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Question Number 201065 by mr W last updated on 28/Nov/23

Commented by mr W last updated on 28/Nov/23

$$findthesidelength\mathit{a}ofthesquare$$$$intermsof\mathit{r}.$$

Answered by ajfour last updated on 29/Nov/23

$$2r\cos \theta =\frac{a}{2}$$$$2r\sin \theta =a+p$$$$p^2+{(r-\frac{a}{2})}^2=r^2$$$$\Rightarrow ar=p^2+\frac{a^2}{4}$$$$4r^2=\frac{a^2}{4}+a^2{(1+\sqrt{\frac{r}{a}-\frac{1}{4}})}^2$$$$say\frac{r}{a}=t$$$$4t^2=\frac{1}{4}+{(1+\sqrt{t-\frac{1}{4}})}^2$$$$4(t-\frac{1}{4})(t+\frac{1}{4})={(1+\sqrt{t-\frac{1}{4}})}^2$$$$sayt-\frac{1}{4}=z^2$$$$4z^2(z^2+\frac{1}{2})=1+z^2+2z$$$$4z^4+z^2-2z-1=0$$$$z\approx 0.83813$$$$t=\frac{1}{4}+z^2\approx 0.95246$$$$a=\frac{r}{t}\approx \frac{r}{0.95246}\approx 1.0499r$$

Commented by mr W last updated on 29/Nov/23

$$thankssir!excellent!$$

Answered by Frix last updated on 29/Nov/23

$$r=1$$$$\sqrt{4-x^2}-\sqrt{2x-x^2}=2x$$$$\mathrm{The}\mathrm{usual}\mathrm{squaring}\mathrm{etc}.\mathrm{leads}\mathrm{to}$$$$x^4-\frac{x^3}{2}-\frac{7x^2}{8}-\frac{x}{2}+\frac{1}{2}=0$$$$\mathrm{No}\mathrm{useable}\mathrm{exact}\mathrm{solution}.$$$$x\approx .524956679452$$$$a=2x\approx 1.0499133589$$

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